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I am in $\mathbb{R}$ with the usual metric and i want to check if $E= [1,\infty)$ which is a subset of $\mathbb{R}$ is connected. I have a theorem that states that if E is connected then for $a,b∈E$ whenever we have $a < x < b $ means that $x∈E$. I believe the solution lays in this theorem but it is not really clear to me just yet.

ps. I am also taught the way of proving a set is non connective by using 2 open sets but i dont think that is of any use here.

Edward Colens
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    Your theorem presumably assumes that $E$ is a subset of $\mathbb{R}$. PS you don't need a metric to define connectedness. – lcv Nov 29 '22 at 16:22
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    A subset $X$ of the Euclidean line is connected if and only if it has the intermediate value property: $$a,b\in X\wedge a\le b\implies[a,b]\subseteq X$$That includes singletons (degenerate intervals). In particular, that applies to your set – FShrike Nov 29 '22 at 16:22

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you could prove that by the statement : [ path connected ] $\implies $ [connected]

all we need is to prove E is path-connected, let any $ a , b ∈ E $ , we should prove that, there is an F which is continues such as $ F(0)=a $ and $ F(b)=b $ .

let $ a<b $ , then we could choose $ F(x) = a + x*d(a,b) $,$ x∈[0,1] $ , where d is the usual metric on E.

F is continues , $ img(F) = [a,b] $ and for every open set A in [a,b] , its easy to see $ F^{-1} (A) $ also open ( for every (x1,x2) open ball in [a,b] , let x2>x1 ,$ F^{-1}((x1,x2)) = (\frac{d(a,x1)}{d(a,b)} ,\frac{d(a,x2)}{d(a,b)} ) $ )

if you prove that every open set in E under the usual metric could be written as sum of open balls as (x1,x2) which I assume you already know, then the $ F^{-1}(A) $=sum of open sets = open set ,and the proof is complete.

ya.sa
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